\(\int \frac {A+B x}{x^{3/2} (b x+c x^2)^{5/2}} \, dx\) [228]

Optimal result

Integrand size = 24, antiderivative size = 213 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 c^2 (b B-A c) x^{3/2}}{3 b^4 \left (b x+c x^2\right )^{3/2}}+\frac {2 c^2 (3 b B-4 A c) \sqrt {x}}{b^5 \sqrt {b x+c x^2}}-\frac {A \sqrt {b x+c x^2}}{3 b^3 x^{7/2}}-\frac {(6 b B-17 A c) \sqrt {b x+c x^2}}{12 b^4 x^{5/2}}+\frac {c (22 b B-41 A c) \sqrt {b x+c x^2}}{8 b^5 x^{3/2}}-\frac {35 c^2 (2 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{11/2}} \] Output:

2/3*c^2*(-A*c+B*b)*x^(3/2)/b^4/(c*x^2+b*x)^(3/2)+2*c^2*(-4*A*c+3*B*b)*x^(1 
/2)/b^5/(c*x^2+b*x)^(1/2)-1/3*A*(c*x^2+b*x)^(1/2)/b^3/x^(7/2)-1/12*(-17*A* 
c+6*B*b)*(c*x^2+b*x)^(1/2)/b^4/x^(5/2)+1/8*c*(-41*A*c+22*B*b)*(c*x^2+b*x)^ 
(1/2)/b^5/x^(3/2)-35/8*c^2*(-3*A*c+2*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2 
)/x^(1/2))/b^(11/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\frac {\sqrt {b} \left (2 b B x \left (-6 b^3+21 b^2 c x+140 b c^2 x^2+105 c^3 x^3\right )-A \left (8 b^4-18 b^3 c x+63 b^2 c^2 x^2+420 b c^3 x^3+315 c^4 x^4\right )\right )+105 c^2 (-2 b B+3 A c) x^3 (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{24 b^{11/2} x^{3/2} (x (b+c x))^{3/2}} \] Input:

Integrate[(A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x]
 

Output:

(Sqrt[b]*(2*b*B*x*(-6*b^3 + 21*b^2*c*x + 140*b*c^2*x^2 + 105*c^3*x^3) - A* 
(8*b^4 - 18*b^3*c*x + 63*b^2*c^2*x^2 + 420*b*c^3*x^3 + 315*c^4*x^4)) + 105 
*c^2*(-2*b*B + 3*A*c)*x^3*(b + c*x)^(3/2)*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/ 
(24*b^(11/2)*x^(3/2)*(x*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1220, 1135, 1132, 1135, 1132, 1136, 220}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(2 b B-3 A c) \int \frac {1}{\sqrt {x} \left (c x^2+b x\right )^{5/2}}dx}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(2 b B-3 A c) \left (-\frac {7 c \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{5/2}}dx}{4 b}-\frac {1}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\right )}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1132

\(\displaystyle \frac {(2 b B-3 A c) \left (-\frac {7 c \left (\frac {5 \int \frac {1}{\sqrt {x} \left (c x^2+b x\right )^{3/2}}dx}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {1}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\right )}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(2 b B-3 A c) \left (-\frac {7 c \left (\frac {5 \left (-\frac {3 c \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {1}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\right )}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1132

\(\displaystyle \frac {(2 b B-3 A c) \left (-\frac {7 c \left (\frac {5 \left (-\frac {3 c \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {1}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\right )}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(2 b B-3 A c) \left (-\frac {7 c \left (\frac {5 \left (-\frac {3 c \left (\frac {2 \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {1}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\right )}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {(2 b B-3 A c) \left (-\frac {7 c \left (\frac {5 \left (-\frac {3 c \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{3 b}+\frac {2 \sqrt {x}}{3 b \left (b x+c x^2\right )^{3/2}}\right )}{4 b}-\frac {1}{2 b \sqrt {x} \left (b x+c x^2\right )^{3/2}}\right )}{2 b}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}\)

Input:

Int[(A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x]
 

Output:

-1/3*A/(b*x^(3/2)*(b*x + c*x^2)^(3/2)) + ((2*b*B - 3*A*c)*(-1/2*1/(b*Sqrt[ 
x]*(b*x + c*x^2)^(3/2)) - (7*c*((2*Sqrt[x])/(3*b*(b*x + c*x^2)^(3/2)) + (5 
*(-(1/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) - (3*c*((2*Sqrt[x])/(b*Sqrt[b*x + c*x 
^2]) - (2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)))/(2*b)))/ 
(3*b)))/(4*b)))/(2*b)
 

Defintions of rubi rules used

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 1132

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 
 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 
 4*a*c)))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ 
[0, m, 1] && IntegerQ[2*p]
 

rule 1135

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* 
c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e)))   Int 
[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I 
ntegerQ[2*p]
 

rule 1136

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x 
_Symbol] :> Simp[2*e   Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + 
 b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 
- b*d*e + a*e^2, 0]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (123 A \,c^{2} x^{2}-66 x^{2} B b c -34 A b c x +12 x B \,b^{2}+8 b^{2} A \right )}{24 b^{5} x^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}-\frac {c^{2} \left (-\frac {2 \left (-64 A c +48 B b \right )}{\sqrt {c x +b}}+\frac {32 b \left (A c -B b \right )}{3 \left (c x +b \right )^{\frac {3}{2}}}-\frac {2 \left (105 A c -70 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{16 b^{5} \sqrt {x \left (c x +b \right )}}\)	\(151\)
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (315 A \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-210 B \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{3} x^{4}+315 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{3} x^{3} \sqrt {c x +b}-315 A \sqrt {b}\, c^{4} x^{4}-210 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b^{2} c^{2} x^{3} \sqrt {c x +b}+210 B \,b^{\frac {3}{2}} c^{3} x^{4}-420 A \,b^{\frac {3}{2}} c^{3} x^{3}+280 B \,b^{\frac {5}{2}} c^{2} x^{3}-63 A \,b^{\frac {5}{2}} c^{2} x^{2}+42 B \,b^{\frac {7}{2}} c \,x^{2}+18 A \,b^{\frac {7}{2}} c x -12 B \,b^{\frac {9}{2}} x -8 A \,b^{\frac {9}{2}}\right )}{24 x^{\frac {7}{2}} \left (c x +b \right )^{2} b^{\frac {11}{2}}}\)	\(234\)

Input:

int((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(c*x+b)*(123*A*c^2*x^2-66*B*b*c*x^2-34*A*b*c*x+12*B*b^2*x+8*A*b^2)/b 
^5/x^(5/2)/(x*(c*x+b))^(1/2)-1/16*c^2/b^5*(-2*(-64*A*c+48*B*b)/(c*x+b)^(1/ 
2)+32/3*b*(A*c-B*b)/(c*x+b)^(3/2)-2*(105*A*c-70*B*b)/b^(1/2)*arctanh((c*x+ 
b)^(1/2)/b^(1/2)))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.25 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\left [-\frac {105 \, {\left ({\left (2 \, B b c^{4} - 3 \, A c^{5}\right )} x^{6} + 2 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{5} + {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (8 \, A b^{5} - 105 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} - 140 \, {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{3} - 21 \, {\left (2 \, B b^{4} c - 3 \, A b^{3} c^{2}\right )} x^{2} + 6 \, {\left (2 \, B b^{5} - 3 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}, \frac {105 \, {\left ({\left (2 \, B b c^{4} - 3 \, A c^{5}\right )} x^{6} + 2 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{5} + {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (8 \, A b^{5} - 105 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} - 140 \, {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{3} - 21 \, {\left (2 \, B b^{4} c - 3 \, A b^{3} c^{2}\right )} x^{2} + 6 \, {\left (2 \, B b^{5} - 3 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}\right ] \] Input:

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
 

Output:

[-1/48*(105*((2*B*b*c^4 - 3*A*c^5)*x^6 + 2*(2*B*b^2*c^3 - 3*A*b*c^4)*x^5 + 
 (2*B*b^3*c^2 - 3*A*b^2*c^3)*x^4)*sqrt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x 
^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(8*A*b^5 - 105*(2*B*b^2*c^3 - 3*A*b*c^ 
4)*x^4 - 140*(2*B*b^3*c^2 - 3*A*b^2*c^3)*x^3 - 21*(2*B*b^4*c - 3*A*b^3*c^2 
)*x^2 + 6*(2*B*b^5 - 3*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c^2*x^6 
 + 2*b^7*c*x^5 + b^8*x^4), 1/24*(105*((2*B*b*c^4 - 3*A*c^5)*x^6 + 2*(2*B*b 
^2*c^3 - 3*A*b*c^4)*x^5 + (2*B*b^3*c^2 - 3*A*b^2*c^3)*x^4)*sqrt(-b)*arctan 
(sqrt(c*x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) - (8*A*b^5 - 105*(2*B*b^2*c^3 - 3 
*A*b*c^4)*x^4 - 140*(2*B*b^3*c^2 - 3*A*b^2*c^3)*x^3 - 21*(2*B*b^4*c - 3*A* 
b^3*c^2)*x^2 + 6*(2*B*b^5 - 3*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6* 
c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)/x**(3/2)/(c*x**2+b*x)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}} x^{\frac {3}{2}}} \,d x } \] Input:

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")
 

Output:

integrate((B*x + A)/((c*x^2 + b*x)^(5/2)*x^(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\frac {35 \, {\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{5}} + \frac {210 \, {\left (c x + b\right )}^{4} B b c^{2} - 560 \, {\left (c x + b\right )}^{3} B b^{2} c^{2} + 462 \, {\left (c x + b\right )}^{2} B b^{3} c^{2} - 96 \, {\left (c x + b\right )} B b^{4} c^{2} - 16 \, B b^{5} c^{2} - 315 \, {\left (c x + b\right )}^{4} A c^{3} + 840 \, {\left (c x + b\right )}^{3} A b c^{3} - 693 \, {\left (c x + b\right )}^{2} A b^{2} c^{3} + 144 \, {\left (c x + b\right )} A b^{3} c^{3} + 16 \, A b^{4} c^{3}}{24 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} - \sqrt {c x + b} b\right )}^{3} b^{5}} \] Input:

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x, algorithm="giac")
 

Output:

35/8*(2*B*b*c^2 - 3*A*c^3)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^5) + 
 1/24*(210*(c*x + b)^4*B*b*c^2 - 560*(c*x + b)^3*B*b^2*c^2 + 462*(c*x + b) 
^2*B*b^3*c^2 - 96*(c*x + b)*B*b^4*c^2 - 16*B*b^5*c^2 - 315*(c*x + b)^4*A*c 
^3 + 840*(c*x + b)^3*A*b*c^3 - 693*(c*x + b)^2*A*b^2*c^3 + 144*(c*x + b)*A 
*b^3*c^3 + 16*A*b^4*c^3)/(((c*x + b)^(3/2) - sqrt(c*x + b)*b)^3*b^5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {A+B\,x}{x^{3/2}\,{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \] Input:

int((A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x)
 

Output:

int((A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.61 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx=\frac {-315 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a b \,c^{3} x^{3}-315 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{4} x^{4}+210 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{3} c^{2} x^{3}+210 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c^{3} x^{4}+315 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a b \,c^{3} x^{3}+315 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{4} x^{4}-210 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{3} c^{2} x^{3}-210 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c^{3} x^{4}-16 a \,b^{5}+36 a \,b^{4} c x -126 a \,b^{3} c^{2} x^{2}-840 a \,b^{2} c^{3} x^{3}-630 a b \,c^{4} x^{4}-24 b^{6} x +84 b^{5} c \,x^{2}+560 b^{4} c^{2} x^{3}+420 b^{3} c^{3} x^{4}}{48 \sqrt {c x +b}\, b^{6} x^{3} \left (c x +b \right )} \] Input:

int((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x)
 

Output:

( - 315*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*b*c**3*x**3 - 
 315*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*c**4*x**4 + 210* 
sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*b**3*c**2*x**3 + 210*sq 
rt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*b**2*c**3*x**4 + 315*sqrt 
(b)*sqrt(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*a*b*c**3*x**3 + 315*sqrt(b) 
*sqrt(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*a*c**4*x**4 - 210*sqrt(b)*sqrt 
(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*b**3*c**2*x**3 - 210*sqrt(b)*sqrt(b 
 + c*x)*log(sqrt(b + c*x) + sqrt(b))*b**2*c**3*x**4 - 16*a*b**5 + 36*a*b** 
4*c*x - 126*a*b**3*c**2*x**2 - 840*a*b**2*c**3*x**3 - 630*a*b*c**4*x**4 - 
24*b**6*x + 84*b**5*c*x**2 + 560*b**4*c**2*x**3 + 420*b**3*c**3*x**4)/(48* 
sqrt(b + c*x)*b**6*x**3*(b + c*x))